Do gauge bosons really transform according to the adjoint representation of the gauge group?

Question

Its commonly said that gauge bosons transform according to the adjoint representation of the corresponding gauge group. For example, for $SU(2)$ the gauge bosons live in the adjoint $3$ dimensional representation and the gluons in the $8$ dimensional adjoint of $SU(3)$.

Nevertheless, they transform according to

$$ A_μ→A′_μ=UA_μU^\dagger−ig(∂_μU)U^\dagger ,$$

which is not the transformation law for some object in the adjoint representation. For example the $W$ bosons transform according to

$$ (W_μ)_i=(W_μ)_i+∂_μa_i(x)+\epsilon_{ijk}a_j(x)(W_μ)_k.$$

Answer 1

A gauge field transforms in the adjoint of the gauge group, but not in the adjoint (or any other) representation of the group of gauge transformations.

In detail:

Let $G$ be the gauge group, and $\mathcal{G} = \{g : \mathcal{M} \to G \vert g \text{ smooth}\}$ the group of all gauge transformations.

A gauge field $A$ is a connection form on a $G$-principal bundle over the spacetime $\mathcal{M}$, which transforms as $$ A \mapsto g^{-1}Ag + g^{-1}\mathrm{d}g$$ for any smooth $g : \mathcal{M} \to G$. If $g$ is constant, i.e. not only an element of $\mathcal{G}$, but of $G$ itself, this obviously reduces to the adjoint action, so $A$ does transform in the adjoint of $G$, but not in the adjoint of $\mathcal{G}$. With respect to $\mathcal{G}$, it does not transform in any proper linear (or projective) representation in the usual sense, but like an element of a Jet bundle.